Applications of Derivatives
Use local approximations, marginal analysis, implicit differentiation and related rates to interpret changing quantities.
What can a local rate tell us?
A derivative is useful when an exact new measurement is unavailable or when two quantities change together. We will use it to approximate changes, assess production decisions, and connect rates through an equation. Each use needs a domain, a time or production scale, and units.
Actual changes and differentials
Increment and differential
For a change \(\Delta x\), the actual output change is \(\Delta y=f(x+\Delta x)-f(x)\). Set \(dx=\Delta x\); the differential is \(dy=f'(x)dx\). The differential is a linear prediction, not generally the exact change.
Compare the curve and its tangent
For \(f(x)=x^2/2\), start at x=1 and use dx=1.
Show worked solution
The exact change is f(2)−f(1)=2−1/2=3/2. Since f′(x)=x, dy=f′(1)·1=1. The finite step produces a difference of 1/2 between the actual and predicted changes.
A smaller step
For \(f(x)=e^{3x}+x^3\) at x=0, compare Δy and dy when dx=0.1.
Hint
Evaluate the exact difference separately from the derivative approximation.
Show worked solution
The exact change is \(e^{0.3}+0.001-1\approx0.3508588\). The derivative is \(3e^{3x}+3x^2\), giving dy=3(0.1)=0.3. The linear approximation underestimates this increase.
Several measured inputs can change at once. To estimate the change of a product, retain the terms linear in the small input changes and track the discarded product of errors.
Two measured sides
A rectangle has area A=lw. Its measured length is 2% high and width 1% high. Estimate the percentage error in area and compare with the exact percentage.
Show worked solution
The first-order estimate is 3%. The exact multiplier is 1.02×1.01=1.0302, an increase of 3.02%. The extra 0.02 percentage points come from multiplying the two small fractional errors.
Approximate a nearby value
Linearization
The linearization at a is \(L(x)=f(a)+f'(a)(x-a)\). Near a, \(f(x)\approx L(x)\); the accuracy depends on the distance and the curvature.
Choose a nearby base input whose function value and derivative are easy to evaluate. Then use the input difference, not the new input itself, as dx.
A square root without a calculator
Approximate \(\sqrt{16.01}\).
Show worked solution
Use f(x)=√x at a=16. Then f(16)=4, f′(16)=1/8 and dx=0.01. The estimate is \(4+(1/8)(0.01)=4.00125\).
Keep in mind
A tangent approximation is local. It is not a replacement for the original function over a large interval.
Predict the direction of error
For √x near x=16, does the tangent prediction lie above or below the curve?
Hint
The square-root function bends downward.
Show worked solution
Its second derivative is negative for x>0, so the curve is concave downward and the tangent lies above it. The approximation is an overestimate.
Relative and percentage error
For a nonzero reference value, relative error is the absolute error divided by the absolute reference value. Percentage error is 100 times the relative error. A differential estimates propagated measurement error when the input uncertainty is small.
Uncertainty in a sphere
A sphere’s measured radius is 10 cm with uncertainty at most 0.02 cm. Estimate the relative volume error.
Show worked solution
V=4πr³/3, so dV=4πr²dr and |dV|/V=3|dr|/r. The estimate is 3(0.02)/10=0.006, or 0.6%. This is a first-order estimate, not an exact rigorous bound on the finite error.
Compare a prediction with the curve
A differential is a prediction
Cost, revenue and profit rates
Economic quantities
C(x) is total cost; p(x) is price per item; \(R(x)=xp(x)\) is revenue; and \(P(x)=R(x)-C(x)\) is profit. Marginal cost, revenue and profit are \(C'(x),R'(x),P'(x)\), measured in money per item.
A price-demand model often lowers the price as sales increase. Revenue must account for that price change on every item, not just multiply the extra sales by the old price.
Build the model before differentiating
A product has p(x)=10−0.001x pesos per item and C(x)=7000+2x pesos. Find marginal revenue and marginal profit.
Show worked solution
At x=2000, marginal revenue is ₱6 per item and marginal profit is ₱4 per item. A nonnegative-price model permits 0≤x≤10,000.
Interpret a negative marginal profit
For this model, interpret P′(6000)=−4.
Hint
This is a rate, not total profit.
Show worked solution
Near 6000 items, increasing production by one item is predicted to reduce total profit by about ₱4. It does not mean total profit equals −₱4.
The next item: exact versus marginal
Next-item approximation
\(C(x+1)-C(x)\) is the exact added cost of moving from x to x+1 items. The derivative \(C'(x)\) approximates this change using a one-item step in a continuous model.
A measurable approximation error
Let \(C(x)=10000+90x-0.05x^2\). Compare the marginal cost at 500 items with the exact cost of the 501st item.
Show worked solution
C′(x)=90−0.1x, so C′(500)=40 dollars per item. The exact increment is 90−0.05(501²−500²)=39.95 dollars. The marginal approximation exceeds it by 5 cents.
Profit from one more bracelet
A bracelet has p(x)=120−x/20 and C(x)=60x+120. Find the exact added profit from the 601st bracelet and compare it with marginal profit at 600.
Hint
Form P(x), then use both a difference and a derivative.
Show worked solution
P(x)=60x−x²/20−120. Thus P′(600)=0, whereas P(601)−P(600)=60−1201/20=−0.05. Zero marginal profit does not imply a zero discrete increment.
Differentiate an equation without solving it
Implicit differentiation
When an equation relates x and y, treat y as a differentiable function of x locally. Differentiate both sides with respect to x, including a factor y′ whenever differentiating a function of y. Then collect the y′ terms.
Both variables change
Find y′ from \(x^2y-x=5y\).
Show worked solution
This formula is valid at points on the curve where the denominator is nonzero.
Logarithms on a curve
Find the tangent slope to \(\ln(xy)=y^2-1\) at (1,1).
Hint
Differentiate xy with the product rule inside the logarithm.
Show worked solution
At (1,1), 1+y′=2y′, so the slope is 1. The point satisfies the original equation and xy>0.
One input can give several tangent points
An implicit curve need not be the graph of one global function. First find every point at the requested x-coordinate, then compute a slope separately at each valid branch.
Two tangent lines
Find the tangents to \(x^2+y^2-xy-7=0\) where x=1.
Show worked solution
Substitution gives y²−y−6=0, so y=3 or −2. Differentiation gives (2y−x)y′=y−2x. At (1,3), y′=1/5; at (1,−2), y′=4/5. The tangents are y−3=(x−1)/5 and y+2=4(x−1)/5.
Keep in mind
If the coefficient of y′ vanishes, do not divide by zero or declare a slope. Examine the equation and local geometry; a vertical tangent or a singular point may occur.
Choose the changing distance carefully
A rising balloon
A balloon rises vertically at 5 m/s. An observer is 300 m horizontally from its launch point. How fast is the observer-to-balloon distance increasing when the height is 400 m?
Show worked solution
Let h be height and s be line-of-sight distance. The fixed horizontal distance gives s²=300²+h². Differentiate: 2ss′=2hh′. At that instant s=500 m, so s′=400·5/500=4 m/s.
Why not 5 m/s?
Explain why the line-of-sight distance in the balloon example changes more slowly than the height.
Hint
Compare the vertical motion with the sloping line of sight.
Show worked solution
Only the component of the vertical velocity along the line of sight increases s. The factor h/s=4/5 makes s′=(4/5)h′.
Combine production and time rates
Production changes through time
Suppose C(x)=90,000+30x and R(x)=300x−x²/30 dollars. At x=6000 items, production level increases at 500 items per week. Find the time rates of cost, revenue and profit.
Show worked solution
C′=30, R′=300−x/15 and P′=270−x/15. At 6000 these are 30, −100 and −130 dollars per item. Multiply by dx/dt=500: dC/dt=15,000, dR/dt=−50,000 and dP/dt=−65,000 dollars per week.
Keep in mind
Check units as a chain: dollars/item × items/week = dollars/week. A negative revenue rate can occur when the price decrease outweighs the extra items sold.
A branch must be specified
Demand x and price p satisfy (x−100)²=4(p+25). At p=0 and dp/dt=2, find dx/dt.
Show worked solution
The price gives x=90 or x=110. Differentiation gives 2(x−100)x′=4p′. Thus x′=−0.4 on the lower branch and +0.4 on the upper branch. Without a specified branch, there is no unique rate. A demand model should state which branch represents the situation.
References
Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen and Christopher J. Stocker, Calculus for Business, Economics, Life Sciences, and Social Sciences, 14th Global Edition, chapters on limits, differentiation and applications. Mark D. Tomenes, Applied Calculus for Business and Economics I, teaching notes and assessment materials (2025–2026).
Further reading: Gilbert Strang and Edwin Herman, Calculus, Volume 1, OpenStax, chapters 2–4.
